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Here's a puzzle from Jojo's Bizarre Adventure Part 6: Stone Ocean. Let me jojog your memory, it's from chapter 673. Any takers?

Edit: Since there was an official answer found, I have marked that one as correct. However, many of the answers posted below were valid and very creative! I especially enjoyed CobaltZorch, Peregrine Rook and humn's answers. Thanks everyone for your participation.

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    $\begingroup$ Why is that patch of grass behind the lady talking? $\endgroup$ – hmakholm left over Monica Jun 27 '16 at 11:21
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    $\begingroup$ That's so sexually ambiguous... $\endgroup$ – beppe9000 Jun 27 '16 at 12:40
  • $\begingroup$ @humn I believe you, but do you have a source? I think the laws for restricting that kind of stuff in Japan is a bit less strict. $\endgroup$ – Max Li Jun 27 '16 at 21:30
  • $\begingroup$ (Stated at first as unsupported fact:) The above comments make sense if the comic book in the panel was a hasty afterthought to pass censors. The alien's right hand must have originally been drawn to, um, hold the additional match. $\endgroup$ – humn Jun 27 '16 at 21:40
  • $\begingroup$ Note that in this comic, the speech bubbles on the right are read first and the ones on the left are read last. $\endgroup$ – Tanner Swett Jul 23 '16 at 1:21

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I found this image on a forum (bottom of which also viewable here):

enter image description here

It appears the official answer is

break the one match into two pieces and make a multiplication sign, nevermind the fact that it would not fit between the 1 and the 2 without moving the 1.

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    $\begingroup$ I think the image suggests using a dot as a multiplication sign, although I don't understand why you couldn't just make it a decimal point instead... $\endgroup$ – Shagnik Jun 27 '16 at 11:55
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    $\begingroup$ Marking this as correct because it's the official solution, although everyone's answers were all just as good and I read them all. Thanks everyone. $\endgroup$ – Max Li Jun 27 '16 at 15:09
  • $\begingroup$ You could break it twice, creating two very short pieces that would fit. $\endgroup$ – Traubenfuchs Jun 28 '16 at 20:19
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    $\begingroup$ I think it is supposed to be a decimal point. Perhaps it was a localization error? $\endgroup$ – AndFisher Jul 17 '17 at 9:51
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After reading the official answer someone else posted shows this isn't the intended solution, but it seems a pretty tidy solution to me.

Just place the match vertically below the 1.

I 2
I

Which reads as 12 = 1.

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It's a tight squeeze, but


looks like 1/2, i.e., $1/2$.

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I thought the answer might be to

Draw a horizontal on the leftmost character to make it a plus sign

But that feels kinda like cheating.

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    $\begingroup$ That was my first thought, inspired by "no minuses". $\endgroup$ – March Ho Jun 27 '16 at 15:18
  • $\begingroup$ That's what I would say, since it makes it exactly 2 $\endgroup$ – Oak Jun 28 '16 at 12:12
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I would just

light my match and use it to burn up either the lone match or the set of three, thus leaving either a symbolic 1 or symbolic 2, which are both at most 2, without using a minus.

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    $\begingroup$ Hey, no lateral-thinking! :P +1 $\endgroup$ – palsch Jun 27 '16 at 16:55
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You could

put it diagonally to make a comma between the two digits, making "1,2". In some countries, the comma is used instead of a decimal point.

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    $\begingroup$ Alternately, push it into the ground so that only the head is visible. $\endgroup$ – Will Jun 27 '16 at 2:54
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Better answer by acushner, who cannot yet make a separate answer to a protected question.
(This is a placeholder until then. Save upvotes for acushner's post.)

  
Sure enough,   |2|   =   2   ≤   2 .


The “$\sf\tiny NO~MINUSES$” restriction sounds suspiciously like a clue or hint.

No minuses?   Would be no problem ...

... as long as pluses are allowed, for +2 ...

     

... except that this turned out to be a virtual duplicate of Max Li's (puzzle's poser's) answer ...

... other than they use the additional match to “draw”
  (Clever—  almost anything is possible!)


Then again, don't actually need to add any matches because ...

... from Jo's perspective the matches already read “2!” ( =   2×1   =   2 )



But if a match just has to be added, it can be appended to get “2!!” ( =   (2!)!   =   (2)!   =   2 )

Added: “!!” in “2!!” could more cleanly be interpreted as double factorial, giving the same result.

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    $\begingroup$ It looks more like she'd see NÌ… to me. $\endgroup$ – Will Jun 27 '16 at 8:45
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    $\begingroup$ I almost went with $\sf I \over \raise-3mu N$ but it would've failed for N=0 $\endgroup$ – humn Jun 27 '16 at 8:46
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Despite the official answer being posted above, I'd like to suggest a possibility that appealed to me.

Put a matchstick immediately to the right to make a Slashed zero "0"

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    $\begingroup$ this same arrangement also makes it look the abs value of 2, which is 2 or less. $\endgroup$ – acushner Jun 28 '16 at 19:52
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    $\begingroup$ Why not make that variation a separate answer, @acushner, it's excellent. (Argh, looks like the system might not let you add answers to protected questions yet. I'll add it to my answer with credit to you. Send me a comment when you can make at your own answer.) $\endgroup$ – humn Jun 29 '16 at 21:38
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Place the matchstick in a upright position between 1 and 2 which will make it look like a decimal, and the number 1.2

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    $\begingroup$ That was suggested over four hours ago. $\endgroup$ – Peregrine Rook Jun 27 '16 at 7:30
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    $\begingroup$ Safe to upvote: this was suggested alright, but only in a comment rather than an answer $\endgroup$ – humn Jun 27 '16 at 18:50
  • $\begingroup$ Did not go through the comments. Apologies for the added redundancy. $\endgroup$ – Shubham Sanghvi Jun 28 '16 at 4:55
  • $\begingroup$ This is my personal favorite solution, and all the better that you figured it out fresh! And this is not at all redundant because pertinent contents of comments belong in answers. $\endgroup$ – humn Jun 29 '16 at 21:15
  • $\begingroup$ hey @humn ,thanks for the support. New around here and not aware of the norms. $\endgroup$ – Shubham Sanghvi Jul 1 '16 at 6:50
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I was going to offer this as a joke answer, but now that I see that Will's answer is considered to be the official solution, I believe that this is just as good:

$\color{black}{\text{Break the match}}$ $\text{into five pieces of approximately equal length,}$ $\text{and}$ $\text{then}$ $\text{carefully}$ $\text{put them down like this:}$ \begin{align}\huge{12}^{^{\Large{1/4}}}\end{align} $\text{i.e., }\Large{\sqrt[4]{12}}\text{,}$ $\color{black}{\text{which evaluates to 1.8612097}\dots}$

Or, as long as we're being light-hearted and thinking laterally,

$\color{black}{\text{Break the match}}$ $\text{into }\textbf{seven }\text{pieces of the appropriate lengths,}$ $\text{and}$ $\text{then}$ $\text{carefully}$ $\text{put them down like this:}$ \begin{align}\Huge{\sqrt[4]{12}}\end{align}

OK, I guess the new match would need to be about four times as long as a normal one for this to work.

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  • $\begingroup$ Regarding your "four times longer" note: fireplace matches are about four times longer, possibly a bit more. $\endgroup$ – Aiken Drum Aug 23 '17 at 13:46
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The question is only about adding one stick. It makes no restriction on rearranging the others.

So, I would do this :

II $\geq$

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    $\begingroup$ I believe that the prohibition against moving preset matches is implied in this sort of puzzle.  After all, if you could move the ones that are already placed, you could do things like I/III and $\square\,$I.  I don't understand your answer: how does II$\ge$ even evaluate to a value? $\endgroup$ – Peregrine Rook Jun 27 '16 at 7:43
  • $\begingroup$ @PeregrineRook $x \leq 2$ is the set of all elements that are two or less. Puzzles are all about thinking laterally and often out of the box includes making unsaid assumptions. The presence of additional innovative solutions is not ground to rejecting one. $\endgroup$ – Saikat Jun 27 '16 at 9:02
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    $\begingroup$ You can create spoilers using >! . Make sure to have a space between the exclamation mark and your answer. $\endgroup$ – Frozn Jun 27 '16 at 9:02
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    $\begingroup$ @user230452: No, that's an expression that is true if x is less than or equal to two and false otherwise. $\{x\in \mathbb{R} : x\leq 2\}$ is the set of all real numbers less than or equal to two. $\endgroup$ – Deusovi♦ Jun 27 '16 at 15:33
  • $\begingroup$ @Deusovi Yes, you are right. I didn't write it in that set notation form because I wanted the puzzle to be playful, not rigorous. $\endgroup$ – Saikat Jun 27 '16 at 15:48

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